Question

If $I_1={\int }_0^{\frac{\pi }{2}}\frac{\sin x-\cos x}{1+\sin x\cos x}dx$, $I_2={\int }_0^{2\pi }{\cos }^{6}xdx$, $I_3={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\sin }^{3}dx$ and $I_4={\int }_0^1\ln (\frac{1}{x}-1)dx$, then

• A. $I_2=I_3=I_4=0,I_1\ne 0$
• B. $I_1=I_2=I_3=0,I_4\ne 0$
• C. $I_1=I_3=I_4=0,I_2\ne 0$
• D. $I_1=I_2=I_4=0,I_3\ne 0$

Answer 1

The correct option is C $I_1=I_3=I_4=0,I_2\ne 0$

$I_1={\int }_0^{\frac{\pi }{2}}\frac{\sin x-\cos x}{1+\sin x\cos x}dx$
$I_1={\int }_0^{\frac{\pi }{2}}\frac{\sin (\frac{\pi }{2}-x)-\cos (\frac{\pi }{2}-x)}{1+\sin (\frac{\pi }{2}-x)\cos (\frac{\pi }{2}-x)}dx$
$={\int }_0^{\frac{\pi }{2}}\frac{\cos x-\sin x}{1+\sin xcosx}dx=-I_1$
or $I_1=0$
$I_3=0$ as ${\sin }^{3}x$ is odd
$I_4={\int }_0^1\ln \left(\frac{1-x}{x}\right)dx$
$={\int }_0^1\ln \left(\frac{1-(1-x)}{1-x}\right)dx$
$={\int }_0^1\ln \frac{x}{1-x}dx=-I_4$
or $I_4=0$
$I_2={\int }_0^{2\pi }{\cos }^{6}xdx=2{\int }_0^{\pi }{\cos }^{6}xdx\ne 0$